Question: Simplify; express your answer in exponential form. Assume $k\neq 0, y\neq 0$. $\dfrac{{(k^{3}y)^{-3}}}{{(k^{-3}y^{-1})^{-2}}}$
Answer: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(k^{3}y)^{-3} = (k^{3})^{-3}(y)^{-3}}$ On the left, we have ${k^{3}}$ to the exponent ${-3}$ . Now ${3 \times -3 = -9}$ , so ${(k^{3})^{-3} = k^{-9}}$ Apply the ideas above to simplify the equation. $\dfrac{{(k^{3}y)^{-3}}}{{(k^{-3}y^{-1})^{-2}}} = \dfrac{{k^{-9}y^{-3}}}{{k^{6}y^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-9}y^{-3}}}{{k^{6}y^{2}}} = \dfrac{{k^{-9}}}{{k^{6}}} \cdot \dfrac{{y^{-3}}}{{y^{2}}} = k^{{-9} - {6}} \cdot y^{{-3} - {2}} = k^{-15}y^{-5}$